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Question-123461

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Question Number 123461 by Algoritm last updated on 25/Nov/20
Answered by Snail last updated on 25/Nov/20
f(x)=(1−x)^(−1) f(1/4)=4/3
$${f}\left({x}\right)=\left(\mathrm{1}−{x}\right)^{−\mathrm{1}} \\ $$$${f}\left(\mathrm{1}/\mathrm{4}\right)=\mathrm{4}/\mathrm{3} \\ $$
Answered by Olaf last updated on 25/Nov/20
Let P_n (x) = Π_(k=0) ^(n−1) (x+k), n≥1 P_n ((1/4)) = Π_(k=0) ^(n−1) ((1/4)+k) P_n ((1/4)) = Π_(k=0) ^(n−1) ((k+4)/4) P_n ((1/4)) = (((n+3)!)/(4^n 3!)) f(x) = 1+Σ_(n=1() ^∞ ((P_n (x))/(n+1)!)) f((1/4)) = 1+(1/6)Σ_(n=1) ^∞ (((n+3)!)/(4^n (n+1)!)) f((1/4)) = 1+(1/6)Σ_(n=1) ^∞ (((n+2)(n+3))/4^n ) f((1/4)) = (3/2)+(8/3)Σ_(n=1) ^∞ ((n(n+1))/4^n ) Let g(x) = (1/(1−x)) = Σ_(n=0) ^∞ x^n g′(x) = (1/((1−x)^2 )) = Σ_(n=1) ^∞ nx^(n−1) x^2 g′(x) = (x^2 /((1−x)^2 )) = Σ_(n=1) ^∞ nx^(n+1) ((2x(1−x)^2 +2x^2 (1−x))/((1−x)^4 )) = Σ_(n=1) ^∞ n(n+1)x^n If x = (1/4) : ((32)/(27)) = Σ_(n=1) ^∞ ((n(n+1))/4^n ) f((1/4)) = (3/2)+(8/3)×((32)/(27)) = ((755)/(162))
$$\mathrm{Let}\:\mathrm{P}_{{n}} \left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}+{k}\right),\:{n}\geqslant\mathrm{1} \\ $$$$\mathrm{P}_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{4}}+{k}\right) \\ $$$$\mathrm{P}_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{{k}+\mathrm{4}}{\mathrm{4}} \\ $$$$\mathrm{P}_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\left({n}+\mathrm{3}\right)!}{\mathrm{4}^{{n}} \mathrm{3}!} \\ $$$${f}\left({x}\right)\:=\:\mathrm{1}+\underset{{n}=\mathrm{1}\left(\right.} {\overset{\infty} {\sum}}\frac{\mathrm{P}_{{n}} \left({x}\right)}{\left.{n}+\mathrm{1}\right)!} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{3}\right)!}{\mathrm{4}^{{n}} \left({n}+\mathrm{1}\right)!} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\mathrm{4}^{{n}} } \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{8}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}^{{n}} } \\ $$$$\mathrm{Let}\:{g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \\ $$$${g}'\left({x}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}−\mathrm{1}} \\ $$$${x}^{\mathrm{2}} {g}'\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}\left({n}+\mathrm{1}\right){x}^{{n}} \\ $$$$\mathrm{If}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\::\:\frac{\mathrm{32}}{\mathrm{27}}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}^{{n}} } \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{8}}{\mathrm{3}}×\frac{\mathrm{32}}{\mathrm{27}}\:=\:\frac{\mathrm{755}}{\mathrm{162}} \\ $$

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